Write the value of current i3 (in mA ) in the circuit shown in the figure.

Question

Write the value of current i3 (in mA ) in the circuit shown in the figure. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yfdkmokocjin3urf.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

5× 103I1=60 ⇒ I1=(60/5 × 103)=12 mA Potential difference across 10 kΩ=240-60=180 V So, I3× 10× 103=180 ⇒ I3=18 mA I2=I3-I1=6 mA

Q. Write the value of current $i_{3}$ (in $m A$ ) in the circuit shown in the figure.

$5 \times 1 0^{3} I_{1} = 60$
$\Rightarrow I_{1} = \frac{60}{5 \times 1 0 ^{3}} = 12 m A$
Potential difference across
$10 k Ω = 240 - 60 = 180 V$
So, $I_{3} \times 10 \times 1 0^{3} = 180$
$\Rightarrow I_{3} = 18 m A$
$I_{2} = I_{3} - I_{1} = 6 m A$

Q. Write the value of current i3​ (in mA ) in the circuit shown in the figure.

5×103I1​=60 ⇒I1​=5×10360​=12mA Potential difference across 10kΩ=240−60=180V So, I3​×10×103=180 ⇒I3​=18mA I2​=I3​−I1​=6mA

Q. Write the value of current $i_{3}$ (in $m A$ ) in the circuit shown in the figure.

$5 \times 1 0^{3} I_{1} = 60$
$\Rightarrow I_{1} = \frac{60}{5 \times 1 0 ^{3}} = 12 m A$
Potential difference across
$10 k Ω = 240 - 60 = 180 V$
So, $I_{3} \times 10 \times 1 0^{3} = 180$
$\Rightarrow I_{3} = 18 m A$
$I_{2} = I_{3} - I_{1} = 6 m A$